Answer = 0.6
Given, Moles of Benzene = 2.0 moles
Moles of Toulene = 2.0 moles
Partial pressure of Benzene= $p_1 =x_1 p_1^0 = \frac{2}{2+2} \times 120\; torr = 0.5 \times 120\; torr$
Partial pressure of Toulene= $p_2 =x_1 p_1^0 = \frac{2}{2+2} \times 80\; torr = 0.5 \times 80\; torr$
According to Dalton’s law of partial pressures, the total pressure over the solution $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
Total pressure $p = 0.5 \times (120+80) = 100\; torr$
At equilibrium, if $x$ be the mole fraction of Benzene in vapour phase,
Partial pressure of Benzene = Total pressure of the solution $\times$ mole fraction of Benzene in vapour pressure
$120 \times 0.5 = p \times x \quad\quad \Rightarrow 120\; torr \times 0.5 = 100\; torr \times x \quad \quad \Rightarrow x = 0.6$